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Chapter 36: Problem 41

An object is at a distance \(d\) to the left of a flat screen. A converging lens with focal length \(f

Short Answer

Expert verified
Two lens positions exist that form an image on the screen, solved by using the lens equation and finding two object distances. The two images will differ in size; one will be larger and inverted, and the other will be smaller and also inverted.

Step by step solution

01

Use the Lens Equation

To find the lens positions that form an image on the screen, use the lens equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \(d_o\) is the object distance, \(d_i\) is the image distance, and \(f\) is the focal length of the lens.
02

Express Image Distance

Since the screen is a fixed distance \(d\) away from the object, we have \(d_o + d_i = d\). Rearrange this equation to express the image distance in terms of the object distance: \(d_i = d - d_o\).
03

Substitute Image Distance into Lens Equation

Substitute \(d_i\) from Step 2 into the lens equation: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d - d_o}\). This will give you a quadratic equation in terms of \(d_o\).
04

Solve the Quadratic Equation

Solve the quadratic equation for \(d_o\) to find the two possible object distances that yield an image on the screen. These solutions will give you the two positions relative to the object where the lens can be placed to form an image on the screen.
05

Analyze the Image Characteristics

To determine how the two images differ from each other, consider the magnification equation \(m = -\frac{d_i}{d_o}\) and analyze the two solutions for \(d_o\). The negative sign in the magnification equation indicates that the image is inverted. If \(m > 1\), the image is larger than the object; if \(m < 1\), it is smaller.
06

Summary of Image Differences

Determine the magnifications for the two \(d_o\) solutions to summarize how the two images differ. A smaller \(d_o\) will have a larger magnification (yielding a larger and possibly inverted image), while a larger \(d_o\) will produce a smaller magnification (yielding a smaller and also inverted image).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lens Equation
The lens equation is a fundamental tool in optics that relates the focal length of a lens, the distance of the object from the lens, and the distance of the image from the lens. It is generally expressed as:
\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. By manipulating this equation, it's possible to determine where a lens must be placed to produce a clear image on a screen or to find out how the image's properties will change with varying distances.
Focal Length
Focal length, denoted by \( f \), is a distance that describes how strongly a lens converges or diverges light. It is the distance from the lens to the point where parallel rays of light converge to a single point after passing through the lens. In the context of a converging lens, like the one mentioned in the exercise, the focal length is positive and less than a quarter of the distance between the object and the screen.

A shorter focal length indicates a stronger lens that bends the rays of light more sharply, leading to a greater degree of magnification for images produced.
Object Distance
Object distance, represented as \( d_o \), is the distance between the object and the lens. This value is crucial when it comes to determining both the location and the properties of the image produced. In our exercise, the value of \( d_o \) is variable as the lens can move between the object and the screen to find the two positions where a clear image is formed on the screen.
Image Distance
Image distance, denoted by \( d_i \), is the distance between the image and the lens. According to the lens equation, it is intrinsically linked to the object distance and the focal length. For a fixed object and screen distance, the image distance can be expressed in terms of the object distance as \( d_i = d - d_o \). This relationship is pivotal in finding the position of the lens that results in an image being cast on the screen.
Magnification
Magnification is the measure of how much larger or smaller the image is compared to the object. The magnification equation is given by:
\[ m = -\frac{d_i}{d_o} \]
In this equation, \( m \) is the magnification factor, \( d_i \) is the image distance, and \( d_o \) is the object distance. The negative sign indicates that the image formed by a single converging lens will be inverted. When \( m > 1 \), the image is larger than the object, and when \( m < 1 \), it's smaller. By analyzing the magnification for different lens placements, we can deduce the characteristics of the images formed at each position.

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Most popular questions from this chapter

An object located 32.0 \(\mathrm{cm}\) in front of a lens forms an image on a screen 8.00 \(\mathrm{cm}\) behind the lens. (a) Find the focal length of the lens. (b) Determine the magnification. (c) Is the lens converging or diverging?

One end of a long glass rod \((n=1.50)\) is formed into a convex surface with a radius of curvature of \(6.00 \mathrm{cm} .\) An object is located in air along the axis of the rod. Find the image positions corresponding to object distances of (a) \(20.0 \mathrm{cm},(\mathrm{b}) 10.0 \mathrm{cm},\) and \((\mathrm{c}) 3.00 \mathrm{cm}\) from the end of the rod.

A cataract-impaired lens in an eye may be surgically removed and replaced by a manufactured lens. The focal length required for the new lens is determined by the lens-to-retina distance, which is measured by a sonar-like device, and by the requirement that the implant provide for correct distant vision. (a) Assuming the distance from lens to retina is \(22.4 \mathrm{mm},\) calculate the power of the implanted lens in diopters. (b) Because no accommodation occurs and the implant allows for correct distant vision, a corrective lens for close work or reading must be used. Assume a reading distance of 33.0 \(\mathrm{cm}\) and calculate the power of the lens in the reading glasses.

A concave spherical mirror has a radius of curvature of \(20.0 \mathrm{cm} .\) Find the location of the image for object distances of \((\mathrm{a}) 40.0 \mathrm{cm},(\mathrm{b}) 20.0 \mathrm{cm},\) and \((\mathrm{c}) 10.0 \mathrm{cm} .\) For each case, state whether the image is real or virtual and upright or inverted. Find the magnification in each case.

In some types of optical spectroscopy, such as photoluminescence and Raman spectroscopy, a laser beam exits from a pupil and is focused on a sample to stimulate electromagnetic radiation from the sample. The focusing lens usually has an antireflective coating preventing any light loss. Assume a \(100-\mathrm{mW}\) laser is located 4.80 \(\mathrm{m}\) from the lens, which has a focal length of \(7.00 \mathrm{cm} .\) (a) How far from the lens should the sample be located so that an image of the laser exit pupil is formed on the surface of the sample? (b) If the diameter of the laser exit pupil is \(5.00 \mathrm{mm},\) what is the diameter of the light spot on the sample? (c) What is the light intensity at the spot?
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